HDU 1115 Lifting the Stone（求多边形重心）

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HDU 1115

大意：给你个n，有n个点，然后给你n个点的坐标，求这n个点形成的多边形的重心的坐标。

struct point{    double x, y;} P[1000010];struct line{    point a, b;} ;double xmult(point p1, point p2, point p){    return (p1.x-p.x)*(p2.y-p.y) - (p2.x-p.x)*(p1.y-p.y);}point intersection(line u,line v){point ret=u.a;double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))/((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));ret.x+=(u.b.x-u.a.x)*t;ret.y+=(u.b.y-u.a.y)*t;return ret;}point barycenter(point a,point b,point c){line u,v;u.a.x=(a.x+b.x)/2;u.a.y=(a.y+b.y)/2;u.b=c;v.a.x=(a.x+c.x)/2;v.a.y=(a.y+c.y)/2;v.b=b;return intersection(u,v);}///多边形重心point Barycenter(int n,point* p){point ret,t;double t1=0,t2;int i;ret.x=ret.y=0;for (i=1;i<n-1;i++)if (fabs(t2=xmult(p[0],p[i],p[i+1]))>eps){t=barycenter(p[0],p[i],p[i+1]);ret.x+=t.x*t2;ret.y+=t.y*t2;t1+=t2;}if (fabs(t1)>eps)ret.x/=t1,ret.y/=t1;return ret;}int T;int n;void Solve(){    scanf("%d", &T);    while(T--)    {        scanf("%d", &n);        for(int i = 0; i < n; ++i)        {            scanf("%lf%lf", &P[i].x, &P[i].y);        }        point t = Barycenter(n, &P[0]);        printf("%.2lf %.2lf\n", t.x+eps, t.y+eps);    }}

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